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Re: Integrating Abs[Sin[]^2]

  • To: mathgroup at smc.vnet.net
  • Subject: [mg38865] Re: [mg38830] Integrating Abs[Sin[]^2]
  • From: Dr Bob <majort at cox-internet.com>
  • Date: Thu, 16 Jan 2003 03:21:10 -0500 (EST)
  • References: <200301150719.CAA23405@smc.vnet.net>
  • Sender: owner-wri-mathgroup at wolfram.com

Integrate[] assumed that k is real, and gave the correct answer for that 
case.  Try this:

Integrate[Abs@Sin[(j + k I) x]^2, {x, 0, 1}]
% /. j -> k
% /. k -> 1

((-k)*Sin[2*j] + j*Sinh[2*k])/  (4*j*k)
((-k)*Sin[2*k] + k*Sinh[2*k])/  (4*k^2)
(1/4)*(-Sin[2] + Sinh[2])

This time, Integrate[] assumed j and k are real, but that makes j + k I 
perfectly general --- except that the answer involves division by the 
imaginary part.  The real version of the answer is the limiting value of 
the complex version:

Integrate[Abs@Sin[(j + k I) x]^2, {x, 0, 1}]
Limit[%, k -> 0]
Integrate[Abs@Sin[j x]^2, {x, 0, 1}]

((-k)*Sin[2*j] + j*Sinh[2*k])/  (4*j*k)
1/2 - Sin[2*j]/(4*j)
(2*j - Sin[2*j])/(4*j)

Bobby

On Wed, 15 Jan 2003 02:19:39 -0500 (EST), Jos R Bergervoet 
<antispam at nospam.com> wrote:

> A strange result appeared when using
>
> Mathematica 4.1 for Linux
> Copyright 1988-2000 Wolfram Research, Inc.
> -- Motif graphics initialized --
>
> in computing the following:
>
> result = Integrate[ Abs[Sin[k x]]^2, {x,0,1}]
> N[ result /. k->I+1 ]
>
> (*  Analytical approach gives 0.261044 + 0.616283 I,  WRONG !!! *)
>
> k=I+1; NIntegrate[ Abs[Sin[k x]^2], {x,0,1}]
>
> (*  Numerical check gives 0.679391  *)
>
>
> So why is the analytical result for |Sin[k x]|^2 wrong?
> What should I do to circumvent such errors?
>
> Thanks in adv.,
>
> -- Jos  <jos.bergervoet at philips.no_s_p_a_m.com>
>
> PS: For those interested, the correct analytical result is:
>
> (Sinh[2Im[k]]/Im[k] - Sin[2Re[k]]/Re[k]) / 4
>
>



-- 
majort at cox-internet.com
Bobby R. Treat



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